Not difficult

Any solution?

luofangxiang wrote: Not difficult JBMO, Of course not difficult.

Ferid.---. wrote: Let $x,y,z,t$ be positive numbers.Prove that $\frac{xyzt}{(x+y)(z+t)}\leq\frac{(x+z)^2(y+t)^2}{4(x+y+z+t)^2}.$ 2017 Azerbaijan JBMO TST ? Thanks.

sqing wrote: Ferid.---. wrote: Let $x,y,z,t$ be positive numbers.Prove that $\frac{xyzt}{(x+y)(z+t)}\leq\frac{(x+z)^2(y+t)^2}{4(x+y+z+t)^2}.$ 2017 Azerbaijan JBMO TST ? Thanks. Yes,can you write your solution?

Easy - we should prove that (1/(1/(x+z)+1/(y+t)))^2/4>=(1/x+1/y)*(1/z+1/t), which is equivalent to: 4*(1/(x+z)+1/(y+t))^2<=(1/x+1/y)(1/z+1/t). 1/(x+z)<=1/2sqrt(xz), 1/(y+t)<=1/2sqrt(yt), and C-S.

Easy one. Here is my solution :(Almost the same with Milos's)

Ferid.---. wrote: Let $x,y,z,t$ be positive numbers.Prove that $\frac{xyzt}{(x+y)(z+t)}\leq\frac{(x+z)^2(y+t)^2}{4(x+y+z+t)^2}.$ Since $$(x+y)(y+t)(t+z)(z+x)-(x+y+z+t)(xyz+xyt+xzt+yzt)=(xt-yz)^2\geq0,$$it's enough to prove that $$(x+z)(y+t)(xyz+xyt+xzt+yzt)\geq4xyzt(x+y+z+t),$$which is $$\sum_{cyc}x^2z(y-t)^2\geq0.$$Done!

Ferid.---. wrote: Let $x,y,z,t$ be positive numbers.Prove that $\frac{xyzt}{(x+y)(z+t)}\leq\frac{(x+z)^2(y+t)^2}{4(x+y+z+t)^2}.$ $ \Leftarrow \ \ \frac{(x+z)^2}{4xz} \cdot \frac{(y+t)^2}{4yt} \ge \frac{(x+y+z+t)^2}{4(x+y)(z+t)} $ $\Leftarrow \ \ ( \frac{(x-z)^2}{4xz}+1)( \frac{(y-t)^2}{4yt}+1) \ge \frac{(x+y+z+t)^2}{4(x+y)(z+t)} $ $\Leftarrow \ \ \frac{(x-z)^2(y-t)^2}{16xyzt} + \frac{(x-z)^2}{4xz}+ \frac{(y-t)^2}{4yt}+1\ge \frac{(x-z+y-t)^2}{4(x+y)(z+t)} +1 $ $\Leftarrow \ \ \frac{(x-z)^2}{4xz}+ \frac{(y-t)^2}{4yt}\ge \frac{(x-z+y-t)^2}{4(x+y)(z+t)} $ $\Leftarrow \ \ \frac{(x-z)^2}{4xz}+ \frac{(y-t)^2}{4yt}\ge \frac{(x-z+y-t)^2}{4xz+4yt} \ge \frac{(x-z+y-t)^2}{4(x+y)(z+t)} $

都科，you do this: $x,y,z\geq 0,x^2+y^2+z^2=3$,prove \[2-x-yz\geq 0\]

mudok wrote: Ferid.---. wrote: Let $x,y,z,t$ be positive numbers.Prove that $\frac{xyzt}{(x+y)(z+t)}\leq\frac{(x+z)^2(y+t)^2}{4(x+y+z+t)^2}.$ $ \Leftarrow \ \ \frac{(x+z)^2}{4xz} \cdot \frac{(y+t)^2}{4yt} \ge \frac{(x+y+z+t)^2}{4(x+y)(z+t)} $ $\Leftarrow \ \ ( \frac{(x-z)^2}{4xz}+1)( \frac{(y-t)^2}{4yt}+1) \ge \frac{(x+y+z+t)^2}{4(x+y)(z+t)} $ $\Leftarrow \ \ \frac{(x-z)^2(y-t)^2}{16xyzt} + \frac{(x-z)^2}{4xz}+ \frac{(y-t)^2}{4yt}+1\ge \frac{(x-z+y-t)^2}{4(x+y)(z+t)} +1 $ $\Leftarrow \ \ \frac{(x-z)^2}{4xz}+ \frac{(y-t)^2}{4yt}\ge \frac{(x-z+y-t)^2}{4(x+y)(z+t)} $ $\Leftarrow \ \ \frac{(x-z)^2}{4xz}+ \frac{(y-t)^2}{4yt}\ge \frac{(x-z+y-t)^2}{4xz+4yt} \ge \frac{(x-z+y-t)^2}{4(x+y)(z+t)} $ My idea is the same as yours

xzlbq wrote: 都科，you do this: $x,y,z\geq 0,x^2+y^2+z^2=3$,prove \[2-x-yz\geq 0\] By Am-Gm, we can easily get $xyz\le 1$. We have $x^2+y^2+z^2+xyz\le 4$. If $x+yz>2$, then $4\ge x^2+xyz+y^2+z^2\ge x^2+xyz+ 2yz=(x+yz)x+2yz>2x+2yz=2(x+yz)>4 $. Contradiction!

mudok wrote: xzlbq wrote: 都科，you do this: $x,y,z\geq 0,x^2+y^2+z^2=3$,prove \[2-x-yz\geq 0\] By Am-Gm, we can easily get $xyz\le 1$. We have $x^2+y^2+z^2+xyz\le 4$. If $x+yz>2$, then $4\ge x^2+xyz+y^2+z^2\ge x^2+xyz+ 2yz=(x+yz)x+2yz>2x+2yz=2(x+yz)>4 $. Contradiction! 反证法，Good!

mudok wrote: xzlbq wrote: 都科，you do this: $x,y,z\geq 0,x^2+y^2+z^2=3$,prove \[2-x-yz\geq 0\] By Am-Gm, we can easily get $xyz\le 1$. We have $x^2+y^2+z^2+xyz\le 4$. If $x+yz>2$, then $4\ge x^2+xyz+y^2+z^2\ge x^2+xyz+ 2yz=(x+yz)x+2yz>2x+2yz=2(x+yz)>4 $. Contradiction! Good method

It must be Belarus national olympiad

muuratjann wrote: It must be Ukraine national olympiad Which year?

luofangxiang wrote: try this: $x,y,z\geq 0,x^2+y^2+z^2=3$,prove \[1+x \geq x(y+z)\] \[x+yz \geq 2xyz\]

Ferid.---. wrote: Let $x,y,z,t$ be positive numbers.Prove that $\frac{xyzt}{(x+y)(z+t)}\leq\frac{(x+z)^2(y+t)^2}{4(x+y+z+t)^2}.$ By computer,we have $$RHS-LHS=\frac{z(z^2y+4zx^2+x^2t+x^2y+3xyz+2zxt+2z^2x+2x^3)(t-y)^2}{8(x+y+z+t)^2(x+y)(z+t)}$$$$+\frac{y(y^2z+2y^2t+4yt^2+t^2x+t^2z+2ytx+3ytz+2t^3)(x-z)^2}{8(x+y+z+t)^2(x+y)(z+t)}$$$$+\frac{(2tx+3xy+3zt)(tx-yz)^2}{8(x+y+z+t)^2(x+y)(z+t)}\ge{0}$$

xzlbq wrote: 都科，you do this: $x,y,z\geq 0,x^2+y^2+z^2=3$,prove \[2-x-yz\geq 0\]

By computer,we have $$2-x-yz=\frac{1}{2}(x-1)^2+\frac{1}{2}(y-z)^2\ge{0}$$

xzlbq wrote: luofangxiang wrote: try this: $x,y,z\geq 0,x^2+y^2+z^2=3$,prove \[1+x \geq x(y+z)\] \[x+yz \geq 2xyz\] By computer,we have $$1+x-x(y+z)=\frac{1}{2}(x+2)(x-1)^2+\frac{1}{2}x(y-1)^2+\frac{1}{2}x(z-1)^2\ge{0}$$ $$x+yz-2xyz=\frac{1}{2}(x-yz)^2+\frac{1}{4}(x+yz)(y-z)^2+\frac{1}{4}(x+yz)(x-1)^2\ge{0}$$

szl6208 wrote: xzlbq wrote: luofangxiang wrote: try this: $x,y,z\geq 0,x^2+y^2+z^2=3$,prove \[1+x \geq x(y+z)\] \[x+yz \geq 2xyz\] By computer,we have $$1+x-x(y+z)=\frac{1}{2}(x+2)(x-1)^2+\frac{1}{2}x(y-1)^2+\frac{1}{2}x(z-1)^2\ge{0}$$ $$x+yz-2xyz=\frac{1}{2}(x-yz)^2+\frac{1}{4}(x+yz)(y-z)^2+\frac{1}{4}(x+yz)(x-1)^2\ge{0}$$ if can not do,tell me, http://www.aoshoo.com/bbs1/dispbbs.asp?boardid=109&Id=25336

OK,I see!

Ferid.---. wrote: muuratjann wrote: It must be Ukraine national olympiad Which year? I am confused. It is Belarussian National Olympiad of this year(9grade)

xzlbq wrote: try this: $x,y,z\geq 0,x^2+y^2+z^2=3$,prove \[x+yz \geq 2xyz\] $x+yz\ge 2\sqrt{xyz}\ge 2xyz$

xzlbq wrote: try this: $x,y,z\geq 0,x^2+y^2+z^2=3$,prove \[1+x \geq x(y+z)\] $x^2+y^2+z^2=3 \Rightarrow \ \ (y+z)^2\le 6-2x^2$ $1+x\ge x(y+z) \Leftarrow (1+x)^2\ge x^2(y+z)^2 \Leftarrow (1+x)^2\ge x^2(6-2x^2) \Leftarrow (x - 1)^2 (2 x^2 + 4 x + 1)\ge 0$

mudok wrote: xzlbq wrote: try this: $x,y,z\geq 0,x^2+y^2+z^2=3$,prove \[1+x \geq x(y+z)\] $x^2+y^2+z^2=3 \Rightarrow \ \ (y+z)^2\le 6-2x^2$ $1+x\ge x(y+z) \Leftarrow (1+x)^2\ge x^2(y+z)^2 \Leftarrow (1+x)^2\ge x^2(6-2x^2) \Leftarrow (x - 1)^2 (2 x^2 + 4 x + 1)\ge 0$ Nice. If $x,y,z\geq 0,x^2+y^2+z^2=3$ . Then $$ x(y+z-1)\leq 1$$

x+y+z<=3

luofangxiang wrote: x+y+z<=3 Of course . $x+y+z\leq \sqrt{3(x^2+y^2+z^2)}$ .

Veri nic solution Mile!

Who's Mile?